Question

What x values is the function concave down if  f(x)=15x⅔+5x ?

Answer 1

Note that f(0)=0

and checking for critical points by taking the derivative and setting to 0

we get

f'(x)=10x-⅓+5=0

or 10x⅓=−5

which simplifies (if x<>0) to

x⅓=−2

→ x=−8

At x=−8

f(−8)=15(−8)23+5(−8)

=15(−2)2+(−40)

=20

Since (−8,20) is the only critical point (other than (0,0) )

and f(x) decreases from x=−8 to x=0

it follows that f(x) decreases on each side of (−8,20), so

f(x) is concave downward when x<0.

When x>0 we simply note that

g(x)=5x is a straight line and

f(x)=15x⅔+5x remains a positive amount (namely 15x⅔above that line

therefore f(x) is not concave downward for x>0.

 

 

 

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f(x)is concave downward when  x<0 .

When  x>0  we simply note that

g(x)=5x is a straight line and

f(x)=15x⅔+5x 

 remains a positive amount (namely  15x⅔  above that line therefore  

f(x)is not concave downward for  x>0